通过考虑 $\left(\frac{2}{x^2} + x^3\right)^{16}$ 的二项式展开式中的通项，解释为什么每一项都与 $x$ 相关。 求 $\left(\frac{2}{x^2} + x^3\right)^{16} (4 - 3x^2)$ 展开式中的常数项（不含 $x$ 的项）。

1. Write down the general term of the binomial expansion

The general term is given by:
$T_{r+1} = \binom{16}{r} \left(\frac{2}{x^2}\right)^{16-r} (x^3)^r$

2. Simplify the general term to isolate the power of $x$

$T_{r+1} = \binom{16}{r} 2^{16-r} x^{-2(16-r)} x^{3r}$
$T_{r+1} = \binom{16}{r} 2^{16-r} x^{-32+2r+3r}$
$T_{r+1} = \binom{16}{r} 2^{16-r} x^{5r-32}$

3. Equate the power of $x$ to zero and explain the contradiction

For a term to be independent of $x$, the power of $x$ must be $0$:
$5r - 32 = 0$
$5r = 32$
$r = 6.4$
Since $r$ must be an integer ($0 \leqslant r \leqslant 16$), $r = 6.4$ is not possible. Therefore, there is no term independent of $x$, which means every term is dependent on $x$.

1. Write down the general term of the binomial expansion

The general term of $\left(\frac{2}{x^2} + x^3\right)^{16}$ is:
$T_{r+1} = \binom{16}{r} \left(2x^{-2}\right)^{16-r} (x^3)^r$
$T_{r+1} = \binom{16}{r} 2^{16-r} x^{-32+2r+3r}$
$T_{r+1} = \binom{16}{r} 2^{16-r} x^{5r-32}$

2. Determine the required powers of $x$ to form a term independent of $x$

To find the term independent of $x$ in $\left(\frac{2}{x^2} + x^3\right)^{16} (4 - 3x^2)$, we need the $x^0$ and $x^{-2}$ terms from the binomial expansion.
For $x^0$: $5r - 32 = 0 \implies r = 6.4$ (reject, as $r$ must be an integer).
For $x^{-2}$: $5r - 32 = -2 \implies 5r = 30 \implies r = 6$.

3. Calculate the term independent of $x$

The $x^{-2}$ term in the expansion is:
$T_7 = \binom{16}{6} 2^{10} x^{-2} = 8008 \times 1024 x^{-2} = 8200192 x^{-2}$
Multiplying by $-3x^2$ from the second bracket:
$\text{Term independent of } x = -3x^2 \times 8200192 x^{-2} = -24600576$